integrationmultivariable-calculus
I found the derivation of finding the Volume of a cylinder using triple integral with Cartesian coordinates. However, I cant seem to find out how they got '2h' in here. Am I missing something?
The derivation can be found in the image below:
$$\iiint_{E}\,dV=\int_{-a}^{a} \int_{-\sqrt{a^2-x^2}}^{\sqrt{a^2-x^2}} \int_0^h\,1 dzdydx=\int_{-a}^{a} 2h \sqrt{a^2-x^2}\,dx=2h\dfrac12\pi a^2=\pi a^2h $$
Thank you so much.
Here is an alternative computation using a single variable integral that confirms your result. The following figure represents the given pyramid.
The equations of the lines situated on the planes $y=0$ and $z=0$ are:
$$y=0,\qquad\frac{x}{a}+\frac{z}{c}=1\Leftrightarrow z=\left( 1-\frac{x}{a}\right) c,$$
$$z=0,\qquad\frac{x}{a}+\frac{y}{c}=1\Leftrightarrow y=\left( 1-\frac{x}{a}\right) b.$$
The intersection of the pyramid with the plane perpendicular to the $x$-axis in $x$ is a right triangle with catheti $\left( 1-\frac{x}{a}\right) c$ and $\left( 1-\frac{x}{a}\right) b$, whose area $A(x)$ is given by
$$A(x)=\frac{1}{2}\left( 1-\frac{x}{a}\right) b\left( 1-\frac{x}{a}\right) c=\frac{bc}{2}\left( 1-\frac{x}{a}\right) ^{2}.$$
Hence the volume is given by the integration of the area $A(x)$ from $x=0$ to $x=a$
$$\begin{eqnarray*} V &=&\int_{0}^{a}A(x)dx \\ &=&\frac{bc}{2}\int_{0}^{a}\left( 1-\frac{x}{a}\right) ^{2}dx \\ &=&\frac{bc}{2}\left( a-\frac{2}{a}\frac{a^{2}}{2}+\frac{1}{a^{2}}\frac{a^{3} }{3}\right) \\ &=&\frac{abc}{6}. \end{eqnarray*}$$
No that is not correct. Both $z$ bounds cannot be constant.
It is much easier in cylindrical coordinates but if you have to set it up in cartesian coordinates,
Integrating wrt $z$ first,
$0 \leq z \leq \sqrt{x^2+y^2-1}$
For $x, y$, the bounds are split.
i) For $0 \leq x \leq 1$, $y$ is bound between two circles in XY-plane ($x^2+y^2 = 1$ and $x^2+y^2 = 10$).
$\sqrt{1-x^2} \leq y \leq \sqrt{10-x^2}$
ii) For $1 \leq x \leq \sqrt{10}$, $y$ is bound between x-axis and circle $x^2+y^2 = 10$.
$0 \leq y \leq \sqrt{10-x^2}$
So the integral would be,
$\displaystyle \int_0^1 \int_{\sqrt{1-x^2}}^{\sqrt{10-x^2}} \int_0^{\sqrt{x^2+y^2-1}} z \ dz \ dy \ dx \ \ + $
$ \displaystyle \int_1^{\sqrt{10}} \int_0^{\sqrt{10-x^2}} \int_0^{\sqrt{x^2+y^2-1}} z \ dz \ dy \ dx$
Best Answer
The steps are the following:$$\int\limits_0^h1dz=h$$and $$\int\limits_{-\sqrt{a^2-x^2}}^{\sqrt{a^2-x^2}}hdy=h\left(y\rvert_{-\sqrt{a^2-x^2}}^{\sqrt{a^2-x^2}}\right)=h\left({\sqrt{a^2-x^2}}-({-\sqrt{a^2-x^2}})\right)=2h\sqrt{a^2-x^2}$$